Pandas DataFrame: copy the contents of a column if it is empty

Question

I have the following DataFrame with named columns and index:

  'a'     'a*'    'b'    'b*'
1  5      NaN     9      NaN
2  NaN    3       3      NaN
3  4      NaN     1      NaN
4  NaN    9       NaN    7

The data source has caused some column headings to be copied slightly differently. For example, as above, some column headings are a string and some are the same string with an additional '*' character.

I want to copy any values (which are not null) from a* and b* columns to a and b, respectively.

Is there an efficient way to do such an operation?

Answer 1

Use np.where

df['a']= np.where(df['a'].isnull(), df['a*'], df['a'])
df['b']= np.where(df['b'].isnull(), df['b*'], df['b'])

Output:

     a  a*  b   b*
0   5.0 NaN 9.0 NaN
1   3.0 3.0 3.0 NaN
2   4.0 NaN 1.0 NaN
3   9.0 9.0 7.0 7.0

Answer 2

Using fillna() is a lot slower than np.where but has the advantage of being pandas only. If you want a faster method and keep it pandas pure, you can use combine_first() which according to the documentation is used to:

Combine Series values, choosing the calling Series’s values first. Result index will be the union of the two indexes

Translation: this is a method designed to do exactly what is asked in the question.

How do I use it?

df['a'].combine_first(df['a*'])

Performance:

df = pd.DataFrame({'A': [0, None, 1, 2, 3, None] * 10000, 'A*': [4, 4, 5, 6, 7, 8] * 10000})

def using_fillna(df):
    return df['A'].fillna(df['A*'])

def using_combine_first(df):
    return df['A'].combine_first(df['A*'])

def using_np_where(df):
    return np.where(df['A'].isnull(), df['A*'], df['A'])

def using_np_where_numpy(df):
    return np.where(np.isnan(df['A'].values), df['A*'].values, df['A'].values)

%timeit -n 100 using_fillna(df)
%timeit -n 100 using_combine_first(df)
%timeit -n 100 using_np_where(df)
%timeit -n 100 using_np_where_numpy(df)

1.34 ms ± 71.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
281 µs ± 15.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
257 µs ± 16.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
166 µs ± 10.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 3

For better performance is possible use numpy.isnan and convert Series to numpy arrays by values:

df['a'] = np.where(np.isnan(df['a'].values), df['a*'].values, df['a'].values)
df['b'] = np.where(np.isnan(df['b'].values), df['b*'].values, df['a'].values)

Another general solution if exist only pairs with/without * in columns of DataFrame and is necessary remove * columns:

First create MultiIndex by split with append *val:

df.columns = (df.columns + '*val').str.split('*', expand=True, n=1)

And then select by DataFrame.xs for DataFrames, so DataFrame.fillna working very nice:

df = df.xs('*val', axis=1, level=1).fillna(df.xs('val', axis=1, level=1))
print (df)
     a    b
1  5.0  9.0
2  3.0  3.0
3  4.0  1.0
4  9.0  7.0

Performance: (depends of number of missing values and length of DataFrame)

df = pd.DataFrame({'A': [0, np.nan, 1, 2, 3, np.nan] * 10000, 
                   'A*': [4, 4, 5, 6, 7, 8] * 10000})

def using_fillna(df):
    df['A'] = df['A'].fillna(df['A*'])
    return df

def using_np_where(df):
    df['B'] = np.where(df['A'].isnull(), df['A*'], df['A'])
    return df

def using_np_where_numpy(df):
    df['C'] = np.where(np.isnan(df['A'].values), df['A*'].values, df['A'].values)
    return df

def using_combine_first(df):
    df['D'] = df['A'].combine_first(df['A*'])
    return df

%timeit -n 100 using_fillna(df)
%timeit -n 100 using_np_where(df)
%timeit -n 100 using_combine_first(df)
%timeit -n 100 using_np_where_numpy(df)

1.15 ms ± 89.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
533 µs ± 13.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
591 µs ± 38.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
423 µs ± 21.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)